Proving convergence of a series. Is my proof correct? Prove that if $\sum_{n=0}^{\infty}{a_{2n}}$ and $\sum_{n=0}^{\infty}{a_{2n+1}}$ are convergent series then $\sum_{n=0}^{\infty}{a_{n}}$ is also convergent
From the assumption we know that $$\forall_{\epsilon>0} \exists_{ N_1>0}\forall_{m\geqslant n\geqslant N_1} |\sum_{k=n}^{m}{a_{2k}}| < \epsilon/2$$
$$\forall_{\epsilon>0} \exists_{ N_2>0}\forall_{m\geqslant n\geqslant N_2} |\sum_{k=n}^{m}{a_{2k+1}}| < \epsilon/2$$
Now let $N'=max\{N_1,N_2\}$ then
$$\forall_{\epsilon>0} \exists_{ N'>0}\forall_{m\geqslant n\geqslant N'}|\sum_{k=n}^{m}{a_{2k}}|+|\sum_{k=n}^{m}{a_{2k+1}}|<\epsilon$$
It's true that
$$|\sum_{k=n}^{m}({a_{2k}+a_{2k+1}})|=|\sum_{k=n}^{m}{a_{2k}}+\sum_{k=n}^{m}{a_{2k}}|\leqslant|\sum_{k=n}^{m}{a_{2k}}|+|\sum_{k=n}^{m}{a_{2k}}|<\epsilon$$
and 
$$|\sum_{k=n}^{m}({a_{2k}+a_{2k+1}})|=|\sum_{k=2n}^{2m+1}{a_{k}}|$$ Therefore, I can take any $\epsilon$ and I'm always able to choose $N'$ so that   
$1)$
$$\forall_{m\geqslant n\geqslant N'}|\sum_{k=2n}^{2m+1}{a_{k}}|<\epsilon$$
I'm not sure if it's enough to say that $\sum_{n=0}^{\infty}{a_{n}}$ converges. The problem is that $1)$ differs a little from the statement:
$2)$
$$\forall_{\epsilon>0} \exists_{ N'>0}\forall_{m'\geqslant n'\geqslant N'} |\sum_{k=n'}^{m'}{a_{k}}|<\epsilon$$ I mean that in $1)$ I can start a sum only with even index whereas in $2)$ I'm not limited. The question is if this limitation makes my proof flawed. Maybe there is a simpler proof for my question?
 A: Let $s_n$ and $t_n$ denote the $n$th partial sums of $\sum_{n = 0}^\infty a_{2n}$ and $\sum_{n = 0}^\infty a_{2n+1}$, respectively. If $r_n$ is the $n$th partial sum of $\sum_{n=0}^\infty a_n$, then $r_{2n} = s_n + t_{n-1}$ and $r_{2n-1} = s_{n-1} + t_{n-1}$, for all $n \ge 1$. Since $\sum_{n = 0}^\infty a_{2n}$ and $\sum_{n = 0}^\infty a_{2n+1}$ are convergent, we can suppose $s_n \to A$ and $t_n \to B$ for some real numbers $A$ and $B$. Then $(r_{2n})_{n=1}^\infty$ and $(r_{2n-1})_{n = 1}^\infty$ both converge to $A + B$. Hence $(r_n)_{n=1}^\infty$ is convergent, which implies $\sum_{n = 0}^\infty a_n$ is convergent.
A: I think you have the right idea, but you are right to be concerned about the fact that in your equation (1), the sum always starts at an even index and ends at an odd index, so it's not quite general.
Instead of starting with
$$\left|\sum_{k=n}^{m}({a_{2k}+a_{2k+1}})\right|$$
(which always starts at the even index $2n$ and ends at the odd index $2m+1$),
consider starting with the more general
$\left|\sum_{k=n}^{m} a_k \right|$, where $n \leq m$ are arbitrary integers greater than $N'$, and splitting this sum into even and odd indices:
$$\begin{align}
\left|\sum_{k=n}^{m} a_k \right| &= \left|\sum_{k=n, k \text{ even}}^{m} a_k + \sum_{k=n, k \text{ odd}}^{m} a_k\right|\\
&\leq \left|\sum_{k=n, k \text{ even}}^{m} a_k \right| + \left|\sum_{k=n, k \text{ odd}}^{m} a_k \right|
\end{align}$$
Both terms on the right hand side are bounded by $\epsilon/2$ as before.
A: Another approach you could use would be to use the sequences of partial sums:
If $\displaystyle T_n=\sum_{k=0}^na_{2k}, \;U_n=\sum_{k=0}^na_{2k+1}, \; \text{and}\;S_n=\sum_{k=0}^na_k$, then $T_n\to T \text{ and }U_n\to U$ for some $T,U\in\mathbb{R}$;
so $S_{2n}=T_n+U_{n-1}\to T+U,\;\;\; S_{2n+1}=T_n+U_n\to T+U$, 
and therefore $S_n\to T+U$.
